Question

If \[{{\lambda }_{Cu}}\] is the wavelength of \[{{K}_{\alpha }}\] X – ray line of copper ( atomic number 29 ) and \[{{\lambda }_{Mo}}\] is the wavelength of \[{{K}_{\alpha }}\] X – ray line of molybdenum ( atomic number 42 ), then the ratio of \[\dfrac{{{\lambda }_{Cu}}}{{{\lambda }_{Mo}}}\] is close to:

Answer 1

Hint: In this question we are asked to calculate the ratio of \[\dfrac{{{\lambda }_{Cu}}}{{{\lambda }_{Mo}}}\] and we are given the wavelength of \[{{K}_{\alpha }}\] X – ray line of copper and molybdenum. Therefore, we will be using The Moseley’s law to solve this question. Moseley’s law gives us the relation between atomic number Z and the wavelength \[\lambda \].

Formula Used:
\[\lambda ={{(Z-1)}^{2}}\]
Where,
\[\lambda \] is the wavelength of the X – ray line
Z is the atomic number

Complete step by step answer:
The relation between wavelength \[\lambda \] and atomic number Z can be obtained from Moseley’s law. Moseley’s law states that the frequency of an atom is directly proportional to the square of the atomic number of the atom. In terms of equation, it can be written as
\[\lambda =a{{(Z-b)}^{2}}\] ………………….. (1)
Where a and b are the constant.
Now, we know that
\[\nu =\dfrac{c}{\lambda }\] ……………….. (2)
Therefore, from (1) and (2) we can say that
\[\lambda ={{(Z-b)}^{-2}}\] …………… (3)
Assuming the constant a = 1
Now for \[{{K}_{\alpha }}\] b is assumed as 1
Therefore, equation (3) becomes
\[\lambda ={{(Z-1)}^{-2}}\] …………… (4)
Now using this above formula for copper
We get,
\[{{\lambda }_{Co}}={{({{Z}_{Co}}-1)}^{-2}}\]
After substituting the given value
We get,
\[{{\lambda }_{Co}}={{(29-1)}^{-2}}\]
Therefore,
\[{{\lambda }_{Co}}=\dfrac{1}{784}\] ………………. (5)
Now applying equation (4) for molybdenum
We get,
\[{{\lambda }_{Mo}}={{({{Z}_{Mo}}-1)}^{-2}}\]
After substituting the atomic number of molybdenum
We get,
\[{{\lambda }_{Mo}}={{(42-1)}^{-2}}\]
Therefore,
\[{{\lambda }_{Mo}}=\dfrac{1}{1681}\] ………………. (6)
Now, we have been asked to calculate \[\dfrac{{{\lambda }_{Cu}}}{{{\lambda }_{Mo}}}\]
From (5) and (6)
We get,
\[\dfrac{{{\lambda }_{Co}}}{{{\lambda }_{Mo}}}=\dfrac{1681}{784}\]
Therefore,
\[\dfrac{{{\lambda }_{Co}}}{{{\lambda }_{Mo}}}=2.144\]
Therefore, the correct answer is, the ratio of \[\dfrac{{{\lambda }_{Cu}}}{{{\lambda }_{Mo}}}\] is 2.144.

Note:
For the K series of X – ray line the value of constant a is \[\sqrt{{}^{3Rc}/{}_{4}}\], where R is the Rydberg constant and c is the speed of light and that of b is 1. Moseley’s law is very significant as it explains the atomic number as a measurable physical quantity. Before Moseley’s experiment, the elements in the periodic table were arranged by increasing atomic weight. However, after his experiment the elements were arranged by the proton count. This explains the placement of elements with less atomic weight before elements with higher atomic mass such as nickel.